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Saxon Calculus Homeschool: Online Textbook Help30 chapters | 247 lessons

Instructor:
*Gerald Lemay*

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

The definition of the derivative involves the operation of taking a limit. In this lesson we look at how to evaluate limits in those cases where the limit problem is a derivative in disguise.

At the masquerade party you want to ask your friend Deriv for the details of his favorite cooking recipe. But Deriv is disguised as the fictitious comic book character Lim. If you can identify Deriv, you will get the information much quicker.

The definition of the derivative involves taking limits. Sometimes, a limit problem is actually a statement of a derivative. In this lesson we look at how to evaluate limits in those cases where the limit problem is a **derivative in disguise**. And just like with Deriv, there are various disguises.

A common way to define the **derivative f '( x)** is with a Δ

Replacing Δ*x* with *h* we get:

Instead of *h*, any letter will do. Like *c*:

These are three ways to express the derivative using limits.

How is a limit evaluated? There are cases where substituting Δ*x* with the **limit value** gives an answer and we are done. Other times, substitution won't work because it gives zero in the denominator. There are other cases where expanding and simplifying the expression will work but this strategy may be very time consuming if the functions are complicated. For example, evaluating:

The limit value is zero. Substituting zero for Δ*x* gives zero in the denominator. However, expanding the numerator and simplifying, allows the Δ*x* in the denominator to cancel:

This function is not too complicated so this approach works. What if 3(*x* + Δ*x*)^2 was 3(*x* + Δ*x*)^5? Simplifying is much more complicated. This lesson is about a better way to find these types of limits by identifying f(*x*). Then, the limit problem answer is f '(*x*).

Really? Let's check. If f(*x*) is 3*x*^2 then replacing *x* with *x* + Δ*x* means f(*x* + Δ*x*) is 3(*x* + Δ*x*)^2. The numerator of our example is f(*x* + Δ*x*) - f(*x*) which is the first of the three definitions for the derivative.

The beauty of all of this is knowing f(*x*).

With f(*x*) identified as 3*x*^2, the derivative is easy. The derivative of 3*x*^2 is 6*x*. Done! The complicated looking limit was a derivative in disguise. The limit evaluates to 6*x*.

Try evaluating:

Compare this to f(*x* + h) - f(*x*). Maybe the f(*x*) is *x*^5 - 2*x*. If this is true, what is f(*x* + h) ?

Replace the *x* with *x* + h. The *x*^5 - 2*x* becomes (*x* + h)^5 - 2(*x* + h) which is (*x* + h)^5 - 2*x* - 2h. See where this is going?

If f(*x*) is *x*^5 - 2*x*, then f(*x* + h) - f(*x*) is (*x* + h)^5 - 2(*x* + h) - *x*^5 + 2*x* which is the numerator of our example.

Okay! f(*x*) is *x*^5 - 2*x*. The derivative of *x*^5 - 2*x* is simply 5*x*^4 - 2. And we are done! Once again, the limit problem was a derivative in disguise. We only have to identify f(*x*) and take the derivative of f(*x*) to evaluate the limit. Oh, Oh. Looks like our friend Deriv has more variations of the Lim disguise.

Sometimes the derivative of f(*x*) is evaluated for some value of *x*. If this value for *x* is *a*, then:

This derivate may be disguised:

The disguises for these problems can be quite good. Our friend Deriv is creative. What if f(*x*) is the square root of *x* ?

If *a* is 16, f(*a*) = f(16) = the square root of 16 = 4. See the 4 in the example?

What about f(*x* + Δ*x*) ? If f(*x*) is the square root of *x*, then f(*x* + Δ*x*) is the square root of *x* + Δ*x*.

For *x* equal to *a*, f(*x* + Δ*x*) is the square root of *a* + Δ*x*.

For *a* equal to 16, f(*a* + Δ*x*) is the square root of 16 + Δ*x*. Do you see how f(*a* + Δ*x*) - f(*a*) is the numerator of our equation for f '(*a*) ?

We have f(*x*) as the square root of *x*. So f '(*x*) is 0.5 (*x*)^-(1/2). And f '(*a*) is 0.5 (*a*)^-(1/2) = 0.5(16)^-(1/2) = 1/8.

There is one more disguise for f '(a):

Try evaluating:

Again, this is one of those limit problems where substituting for *x* gives zero in the denominator. What if f(*x*) is sin(*x*) ? For what value of *x* does sin(*x*) equal 1? Yes, for *x* = 90o, sin(*x*) equals 1. So *a* is 90o which is pi/2.

f(*x*) - f(*a*) is sin(*x*) - sin(pi/2) which equals sin(*x*) - 1 which is the numerator in the limits example. The denominator *x* - *a* is *x* - pi/2. We have our f(*x*).

For f(*x*) = sin(*x*), the derivative of f(*x*) is cos(*x*).

Continuing, if f '(*x*) = cos(*x*) and *a* is pi/2, then f '(*a*) = cos(pi/2) = 0. Done! The limit is the derivative of f(*x*) evaluated at *x* = pi/2. Our limit is zero. Once we see through the disguise and identify the f(*x*), the rest is easy. This actually makes our friend Deriv quite happy.

In some limit problems the **limit value** may be substituted to obtain an answer. In other problems, this is not possible because the denominator will be zero. In some cases the definition of the **derivative f '( x)** is disguised within the limit problem which leads to knowing f(

A related type of limit problem involves the derivative of f(*x*) evaluated at some value *a*. These types of limits may also be derivatives in disguise and once f(*x*) is determined, f '(*x*) is evaluated at *x* = *a*.

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Saxon Calculus Homeschool: Online Textbook Help30 chapters | 247 lessons

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